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| 1. |
What fraction of hydrogen atoms is in the state with the principle quantum number n=2 at a temperature T=3000K? |
| Answer» <html><body><p></p>Solution :By Boltzmann formula <br/> `(N_(2))/(N_(1))=(g_(2))/(g_(1)) e^(DeltaE//kT)`. <br/> Here `DeltaE=` energy difference between `<a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>=1` and `n=2` <a href="https://interviewquestions.tuteehub.com/tag/states-1225920" style="font-weight:bold;" target="_blank" title="Click to know more about STATES">STATES</a> <br/> `=13.6(1-(1)/(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>))eV=10.22eV` <br/> `g_(1)=2` and `g_(2)=8` (counting `2s` & `2P` states).Thus <br/> `(N_(2))/(N_(1))=4e^(-10.22xx1.602xx10^(-19)//1.38xx10^(-23)xx3000)=2.7xx10^(-17)` <br/> Explicity `<a href="https://interviewquestions.tuteehub.com/tag/eta-446786" style="font-weight:bold;" target="_blank" title="Click to know more about ETA">ETA</a>=(N_(2))/(N_(1))=n^(2)e^(-DeltaE//kT),DeltaE_(n)= ħR(1-(1)/(n^(2)))` <br/> for the nth excited state beacuse teh degeneracy of the state with principle quantum number `n is 2n^(2)`.</body></html> | |